1. Preview Section 1 Introduction to Vectors Section 2 Vector Operations
Section 3 Projectile Motion
Section 4 Relative Motion

2. Introduction to Vectors
Scalar - a quantity that has magnitude but no direction
Examples: volume, mass, temperature, speed
Vector - a quantity that has both magnitude and direction
Examples: acceleration, velocity, displacement, force
Emphasize that direction means north, south, east, west, up, or down. It does not mean increasing or decreasing. Even though the temperature may be going “up”, it is really increasing and has no direction. To further emphasize the distinction, point out that it is meaningless to talk about the direction of temperature at a particular point in time, while measurements such as velocity have direction at each moment.

3. Vector Properties Vectors are generally drawn as arrows.
Length represents the magnitude
Arrow shows the direction
Resultant - the sum of two or more vectors

4. Finding the Resultant Graphically
Method
Draw each vector in the proper direction.
Establish a scale (i.e. 1 cm = 2 m) and draw the vector the appropriate length.
Draw the resultant from the tip of the first vector to the tail of the last vector.
Measure the resultant.
The resultant for the addition of a + b is shown to the left as c.
Ask students if a and b have the same magnitude. How can they tell?

5. Vector Addition
Vectors can be moved parallel to themselves without changing the resultant.
the red arrow represents the resultant of the two vectors
Stress that the order in which they are drawn is not important because the resultant will be the same.

6. Vector Addition Vectors can be added in any order.
The resultant (d) is the same in each case
Subtraction is simply the addition of the opposite vector.

7. Sample Resultant Calculation
A toy car moves with a velocity of .80 m/s across a moving walkway that travels at 1.5 m/s. Find the resultant speed of the car.
Use this to demonstrate the graphical method of adding vectors. Use a ruler to measure the two components and determine the scale. Then determine the size and direction of the resultant using the ruler and protractor. This would make a good practice problem for Section 2, when students learn how to add vectors using the Pythagorean theorem and trigonometry.

8. Vector Operations
Use a traditional x-y coordinate system as shown below on the right.
The Pythagorean theorem and tangent function can be used to add vectors.
More accurate and less time-consuming than the graphical method
Direction means north, south, east, west, up, or down. It does not mean increasing or decreasing. So even though the temperature may be going “up,” it is really just increasing and has no direction.

9. Pythagorean Theorem and Tangent Function
Remind students that the Pythagorean theorem can only be used with right triangles.

10. Vector Addition - Sample Problems
12 km east + 9 km east = ?
Resultant: 21 km east
12 km east + 9 km west = ?
Resultant: 3 km east
12 km east + 9 km south = ?
Resultant: 15 km at 37° south of east
12 km east + 8 km north = ?
Resultant: 14 km at 34° north of east
For the first two items, have students predict the answer before showing it. They generally have no trouble with these two problems. Point out that the process is the same if it is km/h or m/s2. Only the units change. These problems do not require trigonometry because the vectors are in the same direction (or opposite directions).
For the third problem, most students will probably remember the Pythagorean theorem and get the magnitude, but many will fail to get the direction or will just write southeast.
Show students how to use the trig identities to determine the angle. Then, explain why it is south of east and not east of south by showing what each direction would look like on an x-y axis. If they draw the 9 km south first and then add the 12 km east, they will get an answer of 53° east of south (which is the same direction as 37° south of east). After your demonstration, have students solve the fourth problem on their own, and then check their answers. Review the solution to this problem also.
Insist that students place arrows on every vector drawn. When they just draw lines, they often draw the resultant in the wrong direction.
You might find the PHet web site helpful ( If you go to the Math simulations, you will find a Vector Addition (flash version). You can download these simulations so your access to the internet is not an issue. You can show both the resultant and components using this simulation. Students could also use this at home to check their solutions to problems.

11. Resolving Vectors Into Components
Review these trigonometry definitions with students to prepare for the next slide (resolving vectors into components).

12. Resolving Vectors into Components
Opposite of vector addition
Vectors are resolved into x and y components
For the vector shown at right, find the vector components vx (velocity in the x direction) and vy (velocity in the y direction). Assume that that the angle is 20.0˚.
Answers:
vx = 89 km/h
vy = 32 km/h
Review the first solution with students, and then let them solve for the second component.

13. Adding Non-Perpendicular Vectors
Four steps
Resolve each vector into x and y components
Add the x components (xtotal = x1 + x2)
Add the y components (ytotal = y1 + y2)
Combine the x and y totals as perpendicular vectors
Explain the four steps using the diagram.
Show students that d1 can be resolved into x1 and y1 . Similarly for d2. Then, the resultant of d1 and d2 (dashed line labeled d) is the same as the resultant of the 4 components.

14. Classroom Practice
A camper walks 4.5 km at 45° north of east and then walks 4.5 km due south. Find the camper’s total displacement.
Answer
3.4 km at 22° S of E
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
Student should note that the 4.5 km south is subtracted from the y component of the first vector that is directed north.

15. Relative Motion Velocity differs in different frames of reference.
Observe as your instructor walks across the front of the room at a steady speed and drops a tennis ball during the walk.
Describe the motion of the ball from the teacher’s frame of reference.
Describe the motion of the ball from a student’s frame of reference.
Which is the correct description of the motion?
Both descriptions are correct. From the teacher’s viewpoint, the ball accelerates straight downward. It stays right below his/her hand and then bounces right back up toward the hand. So the teacher would describe the motion as straight-line acceleration downward. From the student’s perspective, the ball moves sideways as it falls and follows a parabolic path. The horizontal velocity remains constant as it accelerates downward. See the diagram from the text on the next slide.

16. Frames of Reference
A falling object is shown from two different frames of reference:
the pilot (top row)
an observer on the ground (bottom row)
You might want to discuss the old movies of WW I and II pilots after they released the bombs. They always turned to the right or left or sharply upward. This was particularly important if they were flying low. Students should be able to see why in the lower sequence of pictures.

17. Relative Velocity vac = vab + vbc
vac means the velocity of object “a” with respect to frame of reference “c”
Note: vac = -vca
When solving relative velocity problems, follow this technique for writing subscripts.

18. Sample Problem
A boat is traveling downstream. The speed of the boat with respect to Earth (vbe) is 20 km/h. The speed of the river with respect to Earth (vre) is 5 km/h. What is the speed of the boat with respect to the river?
Solution:
vbr = vbe+ ver = vbe + (-vre) = 20 km/h + (-5 km/h)
vbr = 15 km/h
This can more easily be solved using a diagram or common sense, but it is useful to show the students how to manipulate the equation to switch velocities from one frame of reference to another.

19. Classroom Practice Problem
A plane flies northeast at an airspeed of 563 km/h. (Airspeed is the speed of the aircraft relative to the air.) A 48.0 km/h wind is blowing to the southeast. What is the plane’s velocity relative to the ground?
Answer: km/h at 40.1° north of east
How would this pilot need to adjust the direction in order to to maintain a heading of northeast?
You might want to work through Problem F in the text before having students work on this problem.
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

20. Projectile Motion Projectiles: objects that are launched into the air
tennis balls, arrows, baseballs, wrestlers
Gravity affects the motion
Path is parabolic if air resistance is ignored
Path is shortened under the effects of air resistance
Discuss the wide variety of projectiles.
Tell students that the effect of air resistance is significant in many cases, but we will consider ideal examples with gravity being the only force. The effects of air were not very significant in the coin demonstration (see the Notes on the previous slide), but would be much more significant if the objects were traveling faster or had more surface area.
Use the PHET web site to allow students to study projectile motion qualitatively.
Go to simulations, choose “motion,” and choose then choose “projectile motion.”
In this simulation, you can raise or lower the canon. Start with horizontal launches and note that the time in the air is only dependent on the height, and not on the speed of launch. You can change objects, and you can even launch a car. You also have the option of adding air resistance in varying amounts, as well as changing the launch angle.
Have students determine which launch angles produce the same horizontal distance or range (complimentary angles) and find out which launch angle gives the greatest range (45°). Ask them to investigate the effect of air resistance on these results.

21. Components of Projectile Motion
As the runner launches herself (vi), she is moving in the x and y directions.
Remind students that vi is the initial velocity, so it never changes. Students will learn in later slides that vx,i also does not change (there is no acceleration in the horizontal direction) but vy,i does change (because of the acceleration due to gravity).

22. Analysis of Projectile Motion
Horizontal motion
No horizontal acceleration
Horizontal velocity (vx) is constant.
How would the horizontal distance traveled change during successive time intervals of 0.1 s each?
Horizontal motion of a projectile launched at an angle:
Since the initial velocity is constant, the change in x for each successive time interval (such as 0.1 s) will always be the same. Point out that the ball moves the same distance sideways between successive time intervals.
Many students mistakenly believe that the ball is falling straight down eventually. In fact, it keeps moving sideways at a steady rate in the absence of air resistance. With air resistance, it can eventually reach a point where it is falling nearly straight down.

23. Analysis of Projectile Motion
Vertical motion is simple free fall.
Acceleration (ag) is a constant m/s2 .
Vertical velocity changes.
How would the vertical distance traveled change during successive time intervals of 0.1 seconds each?
Vertical motion of a projectile launched at an angle:
Students should note that the vertical distance increases during each successive time interval.
The equations above are simply equations (2), (5), and (4) from the previous section. You might want to write the “old” equations on the board prior to showing them these “new” equations.

24. Projectile Motion - Special Case
Initial velocity is horizontal only (vi,y = 0).
Point out that these equations are the same as those on the previous slides with vi,y = 0 or a launch angle  = 0.
These equations could be used for the coin as it fell off the table (see the Notes on the first slide of this section) or for an object dropped from an airplane flying at a level altitude.
The previous equations (last two slides) are more general and apply to any projectile.

25. Projectile Motion Summary
Projectile motion is free fall with an initial horizontal speed.
Vertical and horizontal motion are independent of each other.
Horizontally the velocity is constant.
Vertically the acceleration is constant (-9.81 m/s2 ).
Components are used to solve for vertical and horizontal quantities.
Time is the same for both vertical and horizontal motion.
Velocity at the peak is purely horizontal (vy = 0).
The 4th and 5th summary points are essential for problem solving. Emphasize these points now, and return to them as students work through problems.

26. Classroom Practice Problem (Horizontal Launch)
People in movies often jump from buildings into pools. If a person jumps horizontally by running straight off a rooftop from a height of 30.0 m to a pool that is 5.0 m from the building, with what initial speed must the person jump?
Answer: 2.0 m/s
As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

27. Classroom Practice Problem (Projectile Launched at an Angle)
A golfer practices driving balls off a cliff and into the water below. The edge of the cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15°, how far does the ball travel horizontally before hitting the water?
Answer: 1.7 x 102 m (170 m)
One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations.
The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.